For which rational degree is its tangent rational?

For which rational number `r in QQ` is its tangent `tan r^o` rational? What about `cos r^o` and `sin r^o`?

We know the obvious ones, `tan 45^o = 1` is rational, is there another rational angle with its tangent rational?

Indeed, there are only all those obvious ones.

If `r in QQ`, then the only rational values `cos r^o` or `sin r^o` can take on are `0, pm 1/2, pm 1`. And the only rational value `tan r^o` can take on is `0` or `pm 1`, when `r` is rational.

Let us prove this. 

We show the case of cosine first.

Recall double-angle formula, where 

`cos(2u) = 2 cos^2(u) -1`

We shall make use of the form `2cos(2u) = (2 cos(u))^2-2`, where we multiply both sides by `2`. 

Let us consider the sequence `c_n = 2 cos(2^n r^o)`, with `c_o = 2 cos(r^o)`. Note this sequence satisfies the relation 

`c_(n+1) = c_n^2 - 2`

Now suppose that `cos(r^o)` is a rational number, then so is `c_0` a rational number, say `c_0 = a/b` where `a,b` some integers in most reduced form, with `gcd(a,b) = 1`.

Then note that `c_1 = a^2/b^2 - 2 = (a^2 - 2b^2)/b^2`.  We claim that `c_1` has denominator `b^2`, namely the expression given is also in most reduced form. Indeed, 

`gcd(a^2 - 2b^2, b^2) = gcd(a^2,b^2) = 1`,

as `gcd(a,b)=1`. Inductively, we see that the denominator of `c_n` is the square of the previous denominator.

Now, if `b ne pm 1`, then the denominator of `c_n` keeps growining, meaning that `c_n` takes on infinitely many different values.

However, if `r = N/M` is a rational number, then `cos(2^n r^o)` can only take on at most `M` many different values, since cosine is `360^o` periodic. This gives a contradiction.

Hence `b` must be `pm 1`.  This means `2 cos(r^o)` is an integer, whence  the only ratoinal values `cos(r^o)` can take on are `0, pm 1/2, pm 1`.

As for sine, note that `sin(r^o) = cos(90^o - r^o)`. So if `r` rational, the only rational values `sin r^o` can take on are also `0, pm 1/2, pm 1`.

Finally, what about tangent?

Note that we can write

cos(2 r^o) = (1 - tan^2(r^o))/(1+ tan^2 (r^o))`

So if `r` rational, and that `tan(r^o) = a/b` also rational, then we must have `cos(2 r^o)` also rational. But since `2r` rational, we must have `cos(2 r^o)` be one of `0, pm 1/2, pm 1`.  Which means `2r` is one of `0, 60, 90, 120, 180, 240, 270, 300, 360` This means `r` is one of `0, 30, 45, 60, 90, 120, 135, 150, 180`. Of these the tangent is rational is only `0, 45, 135`, with value `0, pm 1`.